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No, there does not exist such a $V$.

Let $W = V \cup (V \oplus r)$, and suppose $W = F \Delta Q$. Note that for $s \in {\mathbb Q}$, $W \cap (W \oplus s)$ is nonempty if and only if $s$ or $s-r$ or $s+r$ is an integer. But if $F$ contained an interval of positive length, $W \cap (W \oplus s)$ would be nonempty for all sufficiently small $|s|$. Thus $F$ is nowhere dense, and $W$ is of first category.
But this is impossible, because $[0,1] = \bigcup_{r \in {\mathbb Q}} (W \oplus r)$.

For your second question, the properties of Vitali sets (and finite unions of their translates) that this uses are:

1) there is a dense set of $r$ such that $V \cap (V \oplus r) = \emptyset$.

2) there is a countable set $S$ such that $\bigcup_{r \in S} (V \oplus r) = [0,1)$.

1

No, there does not exist such a $V$.

Let $W = V \cup (V \oplus r)$, and suppose $W = F \Delta Q$. Note that for $s \in {\mathbb Q}$, $W \cap (W \oplus s)$ is nonempty if and only if $s$ or $s-r$ or $s+r$ is an integer. But if $F$ contained an interval of positive length, $W \cap (W \oplus s)$ would be nonempty for all sufficiently small $|s|$. Thus $F$ is nowhere dense, and $W$ is of first category.
But this is impossible, because $[0,1] = \bigcup_{r \in {\mathbb Q}} (W \oplus r)$.